A rod of length 2 text{ cm} makes an angle of | vedclass

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(B) The rod makes an angle of $\frac{2 \pi}{3} 	ext{ rad}$ $(120^{\circ})$ with the principal axis. The angle with the negative x-axis is $60^{\circ}

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The rod makes an angle of $\frac{2 \pi}{3} 	ext{ rad}$ $(120^{\circ})$ with the principal axis. The angle with the negative x-axis is $60^{\circ}$. The horizontal component of the rod is $L_x = 2 \cos(60^{\circ}) = 1 	ext{ cm}$ and the vertical component is $L_y = 2 \sin(60^{\circ}) = \sqrt{3} 	ext{ cm}$.The distance of the object from the lens is $u = -\frac{40}{3} 	ext{ cm}$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} = \frac{1}{10} - \frac{3}{40} = \frac{4-3}{40} = \frac{1}{40}$,so $v = 40 	ext{ cm}$.The lateral magnification is $m = \frac{v}{u} = \frac{40}{-40/3} = -3$.The height of the image is $h_i = |m| \cdot L_y = 3 \cdot \sqrt{3} = 3\sqrt{3} 	ext{ cm}$.The longitudinal magnification is $m_L = -\frac{v^2}{u^2} = -(\frac{40}{-40/3})^2 = -9$.The length of the image along the principal axis is $L'_x = |m_L| \cdot L_x = 9 \cdot 1 = 9 	ext{ cm}$.The angle $\alpha$ made by the image with the principal axis is given by $	an \alpha = \frac{h_i}{L'_x} = \frac{3\sqrt{3}}{9} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.Thus,$\alpha = 30^{\circ} = \frac{\pi}{6} 	ext{ rad}$.Therefore,$n = 6$.

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